Electrons of mass m with de-Broglie wavelength λ fall on the target in an X-ray tube.

⇦

⇨

Electrons of mass m with de-Broglie wavelength λ fall on the target in an X-ray tube. The cut-off wavelength (λ₀) of the emitted X-ray is

Options

(a) λ₀= 2mcλ²/h
(b) λ₀= 2h/mc
(c) λ₀= 2m²c²λ³/h²
(d) λ₀= λ

Correct Answer:

λ₀= 2mcλ²/h

Explanation:

λ = h / p ⇒ p = h / λ

KE of electrons = E = p² / 2m = h² / 2mλ²

Also in X-ray, λ₀ = hc / E = 2mcλ² / h

Related Questions:

1. The energy released by the fission of one uranium atom is 200 MeV.
2. Application of a forward bias to a P-N junction
3. The inductive reactance of an inductor of 1/π henry at 50 Hz frequency is
4. Slope of PV and V for an isobaric process will be
5. Weber ampere per metre is equal to

Topics: Atoms and Nuclei (136)
Subject: Physics (2479)

---

Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score

18000+ students are using NEETLab to improve their score. What about you?

Solve Previous Year MCQs, Mock Tests, Topicwise Practice Tests, Identify Weak Topics, Formula Flash cards and much more is available in NEETLab Android App to improve your NEET score.

---

Share this page with your friends